This is a follow up post after my previous question "Find a function that has finite values for lower norm but becomes infinite"
The conclusion there was that there does not exist a non-decreasing function $h : [0,1) \rightarrow \Re^+$ that satisfies $\|h\|_1=1$ and $\|h\|_a\leq 1$, where $a$ is some value in $[1,2)$, i.e. for lower orders of norm, but has $\|h\|_2 = \infty$.
This was answered based on the continuity of $a \mapsto \|h\|_a$, which implies that there only exists a function $h$ with $\|h\|_2=\infty$ if it satisfies $\|h\|\leq c_a$ where $c_a$ increases in $a$.
My question now is that for any fixed $a$ if it is possible to identify the tightest $c_a$ such that both $\|h\|\leq c_a$ and $\|h\|_2=\infty$ hold for some $h$.
This appears to be very challenging. It is easier to identify some $c_a$ such that both conditions hold, as we may construct a function that satisfies both conditions. But to figure out whether the $c_a$ is the tightest seems much harder. And I have no clue at all how to tackle this. Any thought is much appreciated.
For a fixed function $h$, the optimal constants are precisely $c_a = \|h\|_a$.
If you wish optimal constants $c_a$ such that for any smaller constants $d_a < c_a$ there does not exist a function $f$ such that $\|f\|_a \le d_a$ for $a \in [1,2)$ and $\|f\|_2 = +\infty$, then such a sequence $c_a$ cannot exist.
Indeed, let $c_a$ be a sequence of constants, and $f$ a function such that $\|f\|_a \le c_a$ for $a \in [1,2)$ and $\|f\|_2 = +\infty$. Consider the function $\lambda f$ for any $\lambda \in \langle 0, 1\rangle$.
For $a \in [1,2)$ we have $\|\lambda f\|_a = \lambda\|f\|_a \le \lambda c_a$ and $\|\lambda f\|_2 = \lambda \|f\|_2 = +\infty$.
So, the smaller constants $\lambda c_a$ would work as well.